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andrew-mc [135]
1 year ago
7

a sample of a metal, heated in air, gains mass dm by oxidising at a rate that declines with time t in a parabolic way, meaning t

hat dm2 ¼ kpt where kp is a constant. why does the rate fall off in this way? what does it imply about the nature and utility of the oxide?
Chemistry
1 answer:
velikii [3]1 year ago
5 0

Oxidation originates at the surface of the metal and then it slowly penetrates through the surface.

<em>As at the beginning exposed surface area is more therefore rate of oxidation is also more. After that, the process needed to occur inside the exposed surface of the metal, and the rate of diffusion decreases, so the rate of oxidation decreases.</em>

<em />

<em>It represents oxide layers that reduce the diffusion rate, therefore it is used for producing isolation shields for metals. Apart from that for covering a metal from the external environment or masking oxide layers are produced on metal surfaces.</em>

<em />

<em />

Mass is a quantitative measure of inertia, an essential property of all dependents. It's far, in effect, the resistance that a frame of count number gives to a change in its speed or position upon the application of pressure. The more the mass of a frame, the smaller the alternate produced with the aid of a carried-out force.

Learn more about Mass here:-brainly.com/question/28021242

#SPJ4

<em />

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Part 1: What is the final volume in milliliters when 0.730 L of a 44.8 % (m/v) solution is diluted to 23.3 % (m/v)?
Andre45 [30]

part 1 : the final volume : 1.404 L

part 2 : the initial concentration : 4.06 M

<h3>Further explanation </h3>

Dilution is the process of adding a solvent to get a more dilute solution.

The moles(n) before and after dilution are the same.

Can be formulated :

M₁V₁=M₂V₂

M₁ = Molarity of the solution before dilution  

V₁ = volume of the solution before dilution  

M₂ = Molarity of the solution after dilution  

V₂ = Molarity volume of the solution after dilution

part 1 :

M₁=44.8%

V₁=0.73 L

M₂=23.3%

\tt V_2=\dfrac{M_1.V_1}{M_2}\\\\V_2=\dfrac{44.8\times 0.73}{23.3}\\\\V_2=1.404~L

part 2 :

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